4.2 Quadratics
What is a Quadratic?
A quadratic equation is an equation where the highest power of $x$ is 2. The general form is:
A quadratic equation is an equation where the highest power of $x$ is 2. The general form is:
⚡ General Form:
$$ax^2 + bx + c = 0$$
where $a \neq 0$
• $a$ = coefficient of $x^2$
• $b$ = coefficient of $x$
• $c$ = constant term
• $a$ = coefficient of $x^2$
• $b$ = coefficient of $x$
• $c$ = constant term
💡 Examples of Quadratics:
Three Methods to Solve Quadratics:
$x^2 + 5x + 6 = 0$ (here $a=1$, $b=5$, $c=6$)
$2x^2 - 3x - 2 = 0$ (here $a=2$, $b=-3$, $c=-2$)
$x^2 - 9 = 0$ (here $a=1$, $b=0$, $c=-9$)
$x^2 + 4x = 0$ (here $a=1$, $b=4$, $c=0$)
$2x^2 - 3x - 2 = 0$ (here $a=2$, $b=-3$, $c=-2$)
$x^2 - 9 = 0$ (here $a=1$, $b=0$, $c=-9$)
$x^2 + 4x = 0$ (here $a=1$, $b=4$, $c=0$)
1. Factorising
Quick when it works, but not all quadratics factorise nicely2. Completing the Square
Always works, useful for finding turning points3. Quadratic Formula
Always works, most reliable for any quadratic
🎯 Identify $a$, $b$, and $c$
a =
b =
c =
When $a = 1$: Finding Two Numbers
For $x^2 + bx + c = 0$, we need two numbers that:
For $x^2 + bx + c = 0$, we need two numbers that:
⚡ The Rule:
Find two numbers that:
• Multiply to give $c$
• Add to give $b$
Found via: $x^2 + bx + c = (x + \text{num}_1)(x + \text{num}_2)$
• Multiply to give $c$
• Add to give $b$
Found via: $x^2 + bx + c = (x + \text{num}_1)(x + \text{num}_2)$
Example 1: Factorise $x^2 + 5x + 6$
Step 1: Find two numbers that multiply to 6 and add to 5
Try: $2 \times 3 = 6$ ✓ and $2 + 3 = 5$ ✓
Step 2: Write in factored form
$$x^2 + 5x + 6 = (x + 2)(x + 3)$$
Step 3: Solve $(x + 2)(x + 3) = 0$
$x = -2$ or $x = -3$
Try: $2 \times 3 = 6$ ✓ and $2 + 3 = 5$ ✓
Step 2: Write in factored form
$$x^2 + 5x + 6 = (x + 2)(x + 3)$$
Step 3: Solve $(x + 2)(x + 3) = 0$
$x = -2$ or $x = -3$
Example 2: Factorise $x^2 - 2x - 8$
Step 1: Find two numbers that multiply to $-8$ and add to $-2$
Try: $2 \times (-4) = -8$ ✓ and $2 + (-4) = -2$ ✓
Step 2: Write in factored form
$$x^2 - 2x - 8 = (x + 2)(x - 4)$$
Step 3: Solve
$x = -2$ or $x = 4$
Try: $2 \times (-4) = -8$ ✓ and $2 + (-4) = -2$ ✓
Step 2: Write in factored form
$$x^2 - 2x - 8 = (x + 2)(x - 4)$$
Step 3: Solve
$x = -2$ or $x = 4$
Example 3: Factorise $x^2 - 9x + 20$
Step 1: Find two numbers that multiply to $20$ and add to $-9$
Try: $(-4) \times (-5) = 20$ ✓ and $(-4) + (-5) = -9$ ✓
Answer: $x^2 - 9x + 20 = (x - 4)(x - 5)$
Solutions: $x = 4$ or $x = 5$
Try: $(-4) \times (-5) = 20$ ✓ and $(-4) + (-5) = -9$ ✓
Answer: $x^2 - 9x + 20 = (x - 4)(x - 5)$
Solutions: $x = 4$ or $x = 5$
💡 Sign Guide:
• If $c$ is positive and $b$ is positive: both numbers positive
• If $c$ is positive and $b$ is negative: both numbers negative
• If $c$ is negative: one positive, one negative
• If $c$ is positive and $b$ is negative: both numbers negative
• If $c$ is negative: one positive, one negative
🎯 Factorising Practice
The AC Method
For $ax^2 + bx + c = 0$ where $a \neq 1$, use the AC method:
For $ax^2 + bx + c = 0$ where $a \neq 1$, use the AC method:
⚡ The AC Method:
Step 1: Multiply $a \times c$
Step 2: Find two numbers that multiply to $ac$ and add to $b$
Step 3: Split the middle term using these numbers
Step 4: Factorise by grouping
Step 2: Find two numbers that multiply to $ac$ and add to $b$
Step 3: Split the middle term using these numbers
Step 4: Factorise by grouping
Example 1: Factorise $2x^2 + 5x + 3$
Step 1: $a \times c = 2 \times 3 = 6$
Step 2: Find two numbers that multiply to 6 and add to 5
$2 \times 3 = 6$ ✓ and $2 + 3 = 5$ ✓
Step 3: Split the middle term
$2x^2 + 2x + 3x + 3$
Step 4: Factorise by grouping
$= 2x(x + 1) + 3(x + 1)$
$= (2x + 3)(x + 1)$
Step 2: Find two numbers that multiply to 6 and add to 5
$2 \times 3 = 6$ ✓ and $2 + 3 = 5$ ✓
Step 3: Split the middle term
$2x^2 + 2x + 3x + 3$
Step 4: Factorise by grouping
$= 2x(x + 1) + 3(x + 1)$
$= (2x + 3)(x + 1)$
Example 2: Factorise $3x^2 - 10x + 8$
Step 1: $a \times c = 3 \times 8 = 24$
Step 2: Find two numbers that multiply to 24 and add to $-10$
$(-4) \times (-6) = 24$ ✓ and $(-4) + (-6) = -10$ ✓
Step 3: Split the middle term
$3x^2 - 4x - 6x + 8$
Step 4: Factorise by grouping
$= x(3x - 4) - 2(3x - 4)$
$= (x - 2)(3x - 4)$
Step 2: Find two numbers that multiply to 24 and add to $-10$
$(-4) \times (-6) = 24$ ✓ and $(-4) + (-6) = -10$ ✓
Step 3: Split the middle term
$3x^2 - 4x - 6x + 8$
Step 4: Factorise by grouping
$= x(3x - 4) - 2(3x - 4)$
$= (x - 2)(3x - 4)$
Example 3: Factorise $6x^2 + x - 2$
Step 1: $a \times c = 6 \times (-2) = -12$
Step 2: Find two numbers that multiply to $-12$ and add to $1$
$4 \times (-3) = -12$ ✓ and $4 + (-3) = 1$ ✓
Step 3: Split the middle term
$6x^2 + 4x - 3x - 2$
Step 4: Factorise by grouping
$= 2x(3x + 2) - 1(3x + 2)$
$= (2x - 1)(3x + 2)$
Step 2: Find two numbers that multiply to $-12$ and add to $1$
$4 \times (-3) = -12$ ✓ and $4 + (-3) = 1$ ✓
Step 3: Split the middle term
$6x^2 + 4x - 3x - 2$
Step 4: Factorise by grouping
$= 2x(3x + 2) - 1(3x + 2)$
$= (2x - 1)(3x + 2)$
🎯 AC Method Practice
Number 1:
Number 2:
Difference of Two Squares
When $b = 0$ and $c$ is negative, use DOTS:
When $b = 0$ and $c$ is negative, use DOTS:
Example: Solve $x^2 - 16 = 0$
Taking Out Common Factor First
Using DOTS: $x^2 - 16 = (x + 4)(x - 4) = 0$
Solutions: $x = -4$ or $x = 4$
Solutions: $x = -4$ or $x = 4$
Example: Solve $2x^2 + 6x + 4 = 0$
When $c = 0$ (No Constant)
Step 1: Take out common factor 2
$2(x^2 + 3x + 2) = 0$
Step 2: Factorise the bracket
$2(x + 1)(x + 2) = 0$
Solutions: $x = -1$ or $x = -2$
$2(x^2 + 3x + 2) = 0$
Step 2: Factorise the bracket
$2(x + 1)(x + 2) = 0$
Solutions: $x = -1$ or $x = -2$
Example: Solve $x^2 - 5x = 0$
Step 1: Factor out $x$
$x(x - 5) = 0$
Step 2: Either $x = 0$ or $x - 5 = 0$
Solutions: $x = 0$ or $x = 5$
⚠️ Don't forget $x = 0$ as a solution!
$x(x - 5) = 0$
Step 2: Either $x = 0$ or $x - 5 = 0$
Solutions: $x = 0$ or $x = 5$
⚠️ Don't forget $x = 0$ as a solution!
🎯 Special Case Practice
What is Completing the Square?
Completing the square rewrites $ax^2 + bx + c$ in the form $a(x + p)^2 + q$.
Completing the square rewrites $ax^2 + bx + c$ in the form $a(x + p)^2 + q$.
⚡ The Method (when $a = 1$):
For $x^2 + bx + c$:
Step 1: Halve the coefficient of $x$ to get $p = \frac{b}{2}$
Step 2: Write $(x + p)^2$
Step 3: Subtract $p^2$ and add $c$
Result: $(x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c$
Step 1: Halve the coefficient of $x$ to get $p = \frac{b}{2}$
Step 2: Write $(x + p)^2$
Step 3: Subtract $p^2$ and add $c$
Result: $(x + \frac{b}{2})^2 - (\frac{b}{2})^2 + c$
Example 1: Complete the square for $x^2 + 6x + 5$
Step 1: Halve the coefficient of $x$: $\frac{6}{2} = 3$
Step 2: Write $(x + 3)^2$
But $(x + 3)^2 = x^2 + 6x + 9$, which is 4 too much
Step 3: Adjust: $(x + 3)^2 - 9 + 5 = (x + 3)^2 - 4$
Answer: $x^2 + 6x + 5 = (x + 3)^2 - 4$
Step 2: Write $(x + 3)^2$
But $(x + 3)^2 = x^2 + 6x + 9$, which is 4 too much
Step 3: Adjust: $(x + 3)^2 - 9 + 5 = (x + 3)^2 - 4$
Answer: $x^2 + 6x + 5 = (x + 3)^2 - 4$
Example 2: Complete the square for $x^2 - 4x + 7$
Step 1: Halve the coefficient: $\frac{-4}{2} = -2$
Step 2: Write $(x - 2)^2$
$(x - 2)^2 = x^2 - 4x + 4$
Step 3: Adjust: $(x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$
Answer: $x^2 - 4x + 7 = (x - 2)^2 + 3$
Step 2: Write $(x - 2)^2$
$(x - 2)^2 = x^2 - 4x + 4$
Step 3: Adjust: $(x - 2)^2 - 4 + 7 = (x - 2)^2 + 3$
Answer: $x^2 - 4x + 7 = (x - 2)^2 + 3$
Example 3: Solve $x^2 + 8x + 12 = 0$ by completing the square
Step 1: Complete the square
$(x + 4)^2 - 16 + 12 = 0$
$(x + 4)^2 - 4 = 0$
Step 2: Rearrange
$(x + 4)^2 = 4$
Step 3: Take square root of both sides
$x + 4 = \pm 2$
Step 4: Solve
$x = -4 + 2 = -2$ or $x = -4 - 2 = -6$
$(x + 4)^2 - 16 + 12 = 0$
$(x + 4)^2 - 4 = 0$
Step 2: Rearrange
$(x + 4)^2 = 4$
Step 3: Take square root of both sides
$x + 4 = \pm 2$
Step 4: Solve
$x = -4 + 2 = -2$ or $x = -4 - 2 = -6$
💡 Why Complete the Square?
• Find the vertex (turning point) of a parabola
• The form $(x + p)^2 + q$ tells us the vertex is at $(-p, q)$
• Solve equations that don't factorise easily
• Derive the quadratic formula!
• The form $(x + p)^2 + q$ tells us the vertex is at $(-p, q)$
• Solve equations that don't factorise easily
• Derive the quadratic formula!
🎯 Practice: Complete the Square
Write in the form $(x + p)^2 + q$
Write in the form $(x + p)^2 + q$
(x +
)² +
The Ultimate Solution Method
The quadratic formula works for ANY quadratic equation, even when factorising doesn't work.
The quadratic formula works for ANY quadratic equation, even when factorising doesn't work.
⚡ The Quadratic Formula:
For $ax^2 + bx + c = 0$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Example 1: Solve $x^2 + 5x + 6 = 0$
Here $a = 1$, $b = 5$, $c = 6$
$$x = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2}$$
$x = \frac{-5 + 1}{2} = -2$ or $x = \frac{-5 - 1}{2} = -3$
$$x = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2}$$
$x = \frac{-5 + 1}{2} = -2$ or $x = \frac{-5 - 1}{2} = -3$
Example 2: Solve $2x^2 - 7x + 3 = 0$
Here $a = 2$, $b = -7$, $c = 3$
$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$$ $$x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$$
$x = \frac{7 + 5}{4} = 3$ or $x = \frac{7 - 5}{4} = \frac{1}{2}$
$$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(2)(3)}}{2(2)}$$ $$x = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}$$
$x = \frac{7 + 5}{4} = 3$ or $x = \frac{7 - 5}{4} = \frac{1}{2}$
Example 3: Solve $x^2 + 4x + 1 = 0$ (non-integer answers)
Here $a = 1$, $b = 4$, $c = 1$
$$x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2}$$
$$x = -2 \pm \sqrt{3}$$
$x \approx -0.27$ or $x \approx -3.73$
$$x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = \frac{-4 \pm 2\sqrt{3}}{2}$$
$$x = -2 \pm \sqrt{3}$$
$x \approx -0.27$ or $x \approx -3.73$
💡 The Discriminant: $b^2 - 4ac$
The value under the square root tells us about the solutions:
• $b^2 - 4ac > 0$: Two distinct real solutions
• $b^2 - 4ac = 0$: One repeated real solution
• $b^2 - 4ac < 0$: No real solutions (graph doesn't cross x-axis)
• $b^2 - 4ac > 0$: Two distinct real solutions
• $b^2 - 4ac = 0$: One repeated real solution
• $b^2 - 4ac < 0$: No real solutions (graph doesn't cross x-axis)
🧮 Quadratic Formula Calculator:
x² +
x +
= 0
🎯 What does it mean??
Real Life Uses:
• Physics: Projectile motion (when does the ball hit the ground?)
• Engineering: Calculating dimensions and areas
• Business: Profit maximisation, break-even analysis
• Architecture: Designing arches and parabolic structures
• Physics: Projectile motion (when does the ball hit the ground?)
• Engineering: Calculating dimensions and areas
• Business: Profit maximisation, break-even analysis
• Architecture: Designing arches and parabolic structures
Which Method Should I Use?
• You can spot the numbers easily
• Quick mental calculation
• Sketching the graph
• Converting to vertex form
• Need exact irrational answers
• Complex coefficients
⚡ Decision Guide:
1. First, check for special cases:
• Is there a common factor? Take it out first
• Is $c = 0$? Factor out $x$
• Is it DOTS? Use $(a+b)(a-b)$
2. Try factorising:
• Look for two numbers that multiply to $c$ (or $ac$) and add to $b$
• If you find them quickly, factorise!
3. If factorising is tricky:
• Use the quadratic formula (always works)
• Or complete the square (especially if finding vertex)
• Is there a common factor? Take it out first
• Is $c = 0$? Factor out $x$
• Is it DOTS? Use $(a+b)(a-b)$
2. Try factorising:
• Look for two numbers that multiply to $c$ (or $ac$) and add to $b$
• If you find them quickly, factorise!
3. If factorising is tricky:
• Use the quadratic formula (always works)
• Or complete the square (especially if finding vertex)
✅ Use Factorising when:
• Coefficients are small• You can spot the numbers easily
• Quick mental calculation
✅ Use Completing Square when:
• Finding the turning point• Sketching the graph
• Converting to vertex form
✅ Use Quadratic Formula when:
• Factorising doesn't work• Need exact irrational answers
• Complex coefficients
🎯 Which Method is best?